The derivative is wrong.
Since the t variable is inside the cosine, you would have to use the chain rule to differentiate it.
Chain rule:
Hello,
I am trying to find the value between two maxes/two mins. I just need to see how long a wavelength is.
I have the equation. It is a cos function: 2 + 2.5cos(2pi*60*t).
So, In order to find all maxes and mins, I will derive the above function, and I get:
-sin(2pi*60*t)
I need to set this to zero to find the x values of the max and mins. This is where the problem comes in. I have forgotten how to solve for this:
0 = -sin(2pi*60*t)
I do the inverse of sin of both sides and get this:
0 = -(2pi*60*t)
What is the inverse sin of 0? It's a large range correct?
The next max or min can be found by doing:
pi = -(2pi*60*t) correct?
I tried that, and I get 0 and 1/30, which is def. not the interval between a max and a min. I was going to find the time between a max and a min and multiply by two to find the length of the wavelength. When I calculated this eq. into my graphing calculator, 1/15 is def. not the distance of a wavelength.
Thanks.
You don't really need to know the derivative of "max" or "min" to find the wavelength.
You know (I hope!) that cos(x) has period (wavelength) - that is, one wave starts at x= 0 and ends at . That means that, for this function the wave starts where , or t= 0, and ends where or . This function has a period (wavelength) of .
How are you graphing it? By hand? Graphical calculator? Computer program?
Here's a graph by Wolfram|Alpha of the equation: Wolfram|Alpha, plot[2 + 2.5cos(2pi*60*t)]
Are you graphing it in degrees or radians?